# Can the entropy of an ideal gas change during an isothermal process?

##### 1 Answer

Yes.

#DeltaS_T = nRln(V_2/V_1)# ,i.e. at constant temperature, expanding gases increase in entropy.

Yes, **not** a function of only temperature, so it is **not** zero.

An **isothermal process** has

#dS(T,V) = ((delS)/(delT))_VdT + ((delS)/(delV))_TdV# #" "" "bb((1))#

In this case, one could say that at constant temperature,

#dS_T = ((delS)/(delV))_TdV# #" "" "bb((2.1))#

The **natural variables** associated with this partial derivative are *Helmholtz Maxwell relation*:

#dA = -SdT - PdV# #" "" "bb((3))#

For any state function, the cross-derivatives are equal, so from

#((delS)/(delV))_T = ((delP)/(delT))_V#

Therefore, in terms of a partial derivative that uses the ideal gas law, we get:

#dS_T = ((delP)/(delT))_VdV# #" "" "bb((2.2))#

The right-hand side of

#((delP)/(delT))_V = (del)/(delT)[(nRT)/V]_V = (nR)/V#

Therefore, the **change in entropy of an ideal gas** at a specific, *constant* temperature is (by integrating

#color(blue)(DeltaS_T) = int_((1))^((2)) dS_T = nR int_(V_1)^(V_2) 1/VdV#

#= color(blue)(nRln(V_2/V_1))#

So if the gas expands in the isothermal process, then yes, it will have increased entropy.